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# Stoichiometry example problem 1

## Video transcript

We know that solid phosphorus
will react with chlorine gas to spontaneously produce
phosphorus trichloride, liquid phosphorus trichloride. And we're told that we have 1.45
grams of solid molecular phosphorus. And we're asked how many grams
of chlorine is required to essentially use up all of the
phosphorus that we have? And how many grams of phosphorus
trichloride is going to be produced? Now before you do any of these
stoichiometry problems. And that's just a fancy word for
problems where you need to figure out how much of a certain
reactant is required. Or how much of a product is
going to be produced. Before you do any of these
problems you have to make sure that your reaction, or that your
equation, is balanced. So let's make sure. So on the left-hand side,
here, this molecule of phosphorus has four phosphorus
atoms. So on the whole left-hand side, all of our
reactants combined have 4 phosphorus atoms. So our
products also need to have 4 phosphorus atoms. But the
way it's written right now, I only have 1. So let me just multiply
this guy by 4. Now I have 4 phosphorus atoms on
both sides of the equation. Let's balance the
chlorine now. On the left-hand side I only
have 2 chlorine atoms. This 1 molecule of chlorine
has 2 atoms in it. Here I have each molecule
of phosphorus trichloride has 3 chlorines. And I have 4 molecules of it. So 4 times 3, I have
12 chlorine's on the right-hand side. So 12 on the right-hand side. I need to have 12 on the left. I only have 2 here. Let me multiply that by 6. 6 times 2 is 12. 4 times 3 is 12. Now our equation is
all balanced. 4 phosphoruses each side,
and twelve chlorines. Now the next thing we have to
do, now that we know that we have a balanced equation and
we can get into the meat of the problem, is figure out how
many moles of phosphorus we're dealing with. Because once we know the
moles, we can use stoichiometric ratios. Which is just essentially
saying, look, for every mole of that, I need 6
moles of that. And for every mole of
that, I'm going to produce 4 moles of that. So you want to get it all
in terms of moles. So let's figure out how many
moles of phosphorus we have on our hands. Let's look at our
periodic table. This periodic table, we
have to give proper attribution to the maker. It's made by Le Van
Han Cedric. I got this off of Wikimedia
Creative Commons. It had an attribution license. Other than that, we're
free to use it. And let's go to phosphorus. Phosphorous right here has its
atomic weight of 30.974. Let's just round
that up to 31. Let me write it down
right here. So phosphorus has atomic
weight of 31. Which tells us that a mole of
phosphorus will weigh 31 grams. Remember a mole is this,
you know, 6.02 times 10 to the 20th. It's this huge number of atoms.
If you have that many number of atoms of atomic
phosphorous. It's going to weight at 31
grams. Now if you look at the atomic weight of P4, or a
molecule that has 4 phosphorus atoms in it, it's going
to be 4 times this. So it's going to have an atomic
weight of-- well, what's 4 times 31? it's 124-- of 124. Which means that 1 mole of--
let me write it here-- this tells us, right there, that
1 mole of solid molecular phosphorus is going to have a--
well, since we're doing it in grams, I should say, will
have a mass of 124 grams. Now, given that, we can use that
with this information to figure out how many moles of
molecular phosphorus we have. Solid molecular phosphorus. So let me start over here. So we have 1.45 grams-- let
me write it this way-- of molecular phosphorus, each
molecule containing 4 atoms. And we could just do a little
dimensional analysis, and make sure everything cancels out. Times this information
right here. We have 1 mole of molecular
phosphorus for every 124 grams of molecular phosphorus. I should write this here, just
so we remember what we're talking about. 1 mole of molecular phosphorus
for every 124 grams of molecular phosphorus. And then this cancels out. So the units at least
cancel out. The grams of phosphorus. And then we get 1.45 times
1/124 moles of molecular phosphorus. And we can figure that out. Let's get our calculator out. It's 1.45-- we could just say
divided by 124, right? That's the same thing as
multiplying by 1/124. So divided by 124 is equal to
0.001-- I'll just round it, 7. So this is equal to 0.0117
moles of phosphorus. That's what we're starting
off with. That's what this
1.45 grams are. And that makes sense. If we had an entire mole of
molecular phosphorus, it would weigh a 124 grams. We only
have 1.45 of that. So it's a little bit more than
1/100, which makes sense. This number right here is a
little more than 1/100. Now, we need to think about,
for every mole, let's scroll down here. Although I don't want
to lose my equation. We need to think about for every
mole-- so let me write down, so we have 0.0117 moles
of molecular phosphorus. And for every mole of
phosphorus, how many moles of chlorine molecules do we need? So let me write this down. So for every 6 moles of chlorine
gas-- I'll do it in blue-- for every 6 moles of
chlorine gas, we need 1 mole of molecular phosphorus. And the reason I wrote it this
way, instead of writing 1 mole of molecular phosphorus for
every 6 moles of chlorine gas, is because I want to make sure
the units cancel out. And it also should make sense
for you intuitively. If I have 1 of this, I'm going
to need 6 of this. If I have 0.0117 of this, I'm
going to need 6 times of the chlorine gas. And the units work out. This cancels out with that. And I'm just essentially
multiplying by 6. So we're going to have
6 times 0.0117. I think I still have that
on my calculator. Yes, it's right there. So times 6, is equal to 0.07. I'll just keep it there because
then it is 0 again. So this is equal to 0.07 moles
of chlorine gas required. And I could write
it right here. Let me write, required. Required in the numerator. Required in the denomenator. That makes a little bit more
sense or it clarifies things. For every 6 moles of chlorine
gas that are required, 1 mole of phosphorus is required, of
solid phosphorus, is required. Or we could say for every mole
of phosphorus is required, you need 6 moles of chlorine gas. We got that just from the
original equation. So I'll write this, required,
down right there. So we're almost there. We figured out that 0.07 moles
of chlorine gas are required. But we want to find out
how many grams of chlorine gas are required. So we just have to figure
out how many grams there are per mole? So let's figure that out. Let's look at the chlorine. Chlorine is off here
to the right. This is a super huge
periodic table. Chlorine, right here, has an
atomic weight of 35.453. So chlorine, 35.453
atomic weight. The weighted average of
all of the isotopes of chlorine on earth. So molecular chlorine gas, which
has 2 atoms, is going to have twice that. So I could do it in my head. It is going to be, right,
about what? 70.9? I'll just use a calculator. I don't want to make a
silly math mistake. 35.453 times 2 is
equal to 70.906. I didn't even need
the calculator. So it's equal to 70.906. My memory is bad. 70.906 atomic weight. Which tells us that-- well, let
me write this information that we had down before. We were required to have 0.07
moles of chlorine gas. That is what is required. And let's multiply it. We want moles in the
denominator. So we want moles in the
denominator here. So 1 mole of chlorine gas. For every mole, what's
the mass going to be? Well if the atomic weight of
chlorine gas is 70.906, that means one mole of it is going
to have a mass of 70.906 grams. So for every mole,
we have 70.90 grams. We have 0.07 moles. So we will multiply 0.07 times
70 to figure out how many grams we have. And the
units cancel out. We have moles of chlorine gas
and we're just going to have grams required. So this is going to
be equal to 70 times 0.07 grams required. So let's get that over here. So we already have the 70.906
times 0.07 is equal to 4.96 grams. So 4.96 grams of-- and
actually this is grams of chlorine gas. I should write that there. So grams of chlorine
gas are required. Even the required worked out
with the dimensional analysis. And we've answered the first
part of the problem. If we have 1.45 grams of
phosphorus as one of the reactants, then we're
going to need 4.96. 4.96 grams of chlorine
are required. Now, the second question
is how much phosphorus trichloride is produced? Now there's two ways to do it. An easy way and a hard way. The easy way says, well,
mass is conserved. If you have grams on the side,
the total mass on this side has to be equal to the total
mass on that side. That's the easy way. The slightly harder way is you
could have done the exact same thing we did with the chlorine
gas, you could do it with the phosphorus trichloride. You could say, hey, for
every mole of this, I need 4 moles of that. And then figure out the mass of
each mole and multiply and do all of that. But let's do it the easy way. We know that we have 1.45 grams
of this reactant, and we have 4.96 grams of
this reactant. So this, which is the only
product right here, this product right here must have
a mass of our combined reactants, because it's
going to react fully. So let's figure that out. We have 4.96. I'll just round 4.96 plus 1.45
is equal to 6.41 grams of phosphorus trichloride. 6.41 grams. And we're done.